Problem: Simplify and expand the following expression: $ \dfrac{1}{3a + 24}- \dfrac{3}{2a + 14}- \dfrac{2}{a^2 + 15a + 56} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{1}{3a + 24} = \dfrac{1}{3(a + 8)}$ We can factor a $2$ out of denominator in the second term: $ \dfrac{3}{2a + 14} = \dfrac{3}{2(a + 7)}$ We can factor the quadratic in the third term: $ \dfrac{2}{a^2 + 15a + 56} = \dfrac{2}{(a + 8)(a + 7)}$ Now we have: $ \dfrac{1}{3(a + 8)}- \dfrac{3}{2(a + 7)}- \dfrac{2}{(a + 8)(a + 7)} $ The least common multiple of the denominators is: $ 6(a + 8)(a + 7)$ In order to get the first term over $6(a + 8)(a + 7)$ , multiply by $\dfrac{2(a + 7)}{2(a + 7)}$ $ \dfrac{1}{3(a + 8)} \times \dfrac{2(a + 7)}{2(a + 7)} = \dfrac{2(a + 7)}{6(a + 8)(a + 7)} $ In order to get the second term over $6(a + 8)(a + 7)$ , multiply by $\dfrac{3(a + 8)}{3(a + 8)}$ $ \dfrac{3}{2(a + 7)} \times \dfrac{3(a + 8)}{3(a + 8)} = \dfrac{9(a + 8)}{6(a + 8)(a + 7)} $ In order to get the third term over $6(a + 8)(a + 7)$ , multiply by $\dfrac{6}{6}$ $ \dfrac{2}{(a + 8)(a + 7)} \times \dfrac{6}{6} = \dfrac{12}{6(a + 8)(a + 7)} $ Now we have: $ \dfrac{2(a + 7)}{6(a + 8)(a + 7)} - \dfrac{9(a + 8)}{6(a + 8)(a + 7)} - \dfrac{12}{6(a + 8)(a + 7)} $ $ = \dfrac{ 2(a + 7) - 9(a + 8) - 12} {6(a + 8)(a + 7)} $ Expand: $ = \dfrac{2a + 14 - 9a - 72 - 12}{6a^2 + 90a + 336} $ $ = \dfrac{-7a - 70}{6a^2 + 90a + 336}$